Exercise 9.6 - Chapter 9 Limits and Continuity 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex 9.6
Choose the correct or the most suitable answer from the given four alternatives
Question 1.
$
\lim _{x \rightarrow \infty} \frac{\sin x}{x} \text {.................. }
$
(a) 1
(b) 0
(c) $\infty$
(d) $-\infty$
Solution:
(b) 0
Hint: $\quad \lim _{x \rightarrow \infty} \frac{\sin x}{x}=\lim _{x \rightarrow \infty}\left(\frac{1}{x}\right) \lim _{x \rightarrow \infty}(\sin x)=0$
Question 2.
$
\lim _{x \rightarrow \pi / 2} \frac{(2 x-\pi)}{\cos x}
$
(a) 2
(b) 1
(c) -2
(d) 0
Solution:

(c) -2
$
\begin{aligned}
\text { let } y=2 x-\pi \Rightarrow \frac{\pi+y}{2} & =x \\
x \rightarrow \pi / 2 & \Rightarrow y \rightarrow 0 \\
\therefore \quad \lim _{x \rightarrow \pi / 2}\left(\frac{2 x-\pi}{\cos x}\right) & =\lim _{y \rightarrow 0} \frac{y}{\cos \left(\frac{\pi}{2}+y / 2\right)} \\
& =\lim _{y \rightarrow 0} \frac{y}{-\sin y / 2} \\
& =\lim _{y \rightarrow 0} \frac{-y / 2}{\sin y / 2} \\
& =-2 \lim _{y \rightarrow 0} \frac{1}{\left(\frac{\sin y / 2}{y / 2}\right)}=-2
\end{aligned}
$
Question 3.
$
\lim _{x \rightarrow 0} \frac{\sqrt{1-\cos 2 x}}{x}
$
(a) 0
(b) 1
(c) 2
(d) does not exist
Solution:
(d) does not exist
Hint: $\lim _{x \rightarrow 0} \frac{\sqrt{1-\cos 2 x}}{x}=\lim _{x \rightarrow 0} \frac{\sqrt{1-\cos ^2 x}}{x}=\lim _{x \rightarrow 0} \sqrt{2}\left(\left|\frac{\sin x}{x}\right|\right)=\lim _{x \rightarrow 0^{+}} \frac{\sqrt{2}|\sin x|}{x}=1$ and $\quad \lim _{x \rightarrow 0^{-}} \frac{\sqrt{2}|\sin x|}{x}=-1$
limit does not end
Question 4.
$
\lim _{\theta \rightarrow 0} \frac{\sin \sqrt{\theta}}{\sqrt{\sin \theta}}
$
(a) 1
(b) -1

(c) 0
(d) 2
Solution:
(a) 1
$
\text { Hint: } \quad \begin{aligned}
\lim _{\theta \rightarrow 0} \frac{\sin \sqrt{\theta}}{\sqrt{\sin \theta}} & =\lim _{\theta \rightarrow 0} \frac{\sin \sqrt{\theta}}{\sqrt{\theta}} \cdot \frac{\sqrt{\theta}}{\sqrt{\sin \theta}} \\
& =\lim _{\theta \rightarrow 0} \frac{\sin \sqrt{\theta}}{\sqrt{\theta}} \cdot \lim _{\theta \rightarrow 0} \sqrt{\frac{\theta}{\sin \theta}} \\
& =1
\end{aligned}
$
Question 5.
$\lim _{x \rightarrow \infty}\left(\frac{x^2+5 x+3}{x^2+x+3}\right)^x$ is
(a) $\mathrm{e}^4$
(b) $\mathrm{e}^2$
(c) $\mathrm{e}^3$
(d) 1
Solution:
(a) $\mathrm{e}^4$
$
\begin{aligned}
\lim _{x \rightarrow \infty}\left(\frac{x^2+5 x+3}{x^2+x+3}\right)^x & =\lim _{x \rightarrow \infty}\left(1+\frac{4 x}{x^2+x+3}\right) \\
& =\lim _{x \rightarrow \infty}\left[1+\left(\frac{4 x}{x^2+x+3}\right)\right]^{\frac{x^2+x+3}{4 x} \times \frac{4 x^2}{x^2+x+3}} \\
& =\lim _{x \rightarrow \infty}\left\{\left(1+\frac{4 x}{x^2+x+3}\right)^{\frac{x^2+x+3}{4 x}}\right\}^{\frac{4 x^2}{x^2\left(1+\frac{1}{x}+3 / x^2\right)}} \\
& =e^4
\end{aligned}
$

Question 6.
$\lim _{x \rightarrow \infty} \frac{\sqrt{x^2-1}}{2 x+1}$
(a) 1
(b) 0
(c) -1
(d) $\frac{1}{2}$
Solution:
(d) $\frac{1}{2}$
Hint: $\quad \lim _{x \rightarrow \infty} \frac{\sqrt{x^2-1}}{2 x+1}=\lim _{x \rightarrow 0} \frac{x \sqrt{1-1 / x}}{x(2+1 / x)}=1 / 2$
Question 7.
$\lim _{\boldsymbol{x} \rightarrow \infty} \frac{a^{\boldsymbol{x}}-b^{\boldsymbol{x}}}{x}$
(a) $\log a b$
(b) $\log \left(\frac{a}{b}\right)$
(c) $\log \left(\frac{b}{a}\right)$
(d) $\frac{a}{b}$
Solution:
(b) $\log \left(\frac{a}{b}\right)$
Hint: $\quad \lim _{x \rightarrow \infty} \frac{a^x-b^x}{x}=\lim _{x \rightarrow 0} \frac{a^x-1-b^x+1}{x}$

$
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{\left(a^x-1\right)-\left(b^x+1\right)}{x} \\
& =\lim _{x \rightarrow 0}\left(\frac{a^x-1}{x}\right)-\lim _{x \rightarrow 0}\left(\frac{b^x-1}{x}\right) \\
& =\log a-\log b \\
& =\log (a / b)
\end{aligned}
$
Question 8 .
$
\lim _{x \rightarrow 0} \frac{8^x-4^x-2^x+1^x}{x^2}
$
(a) $2 \log 2$
(b) $2(\log 2)^2$
(c) $\log 2$
(d) $3 \log 2$
Solution:

(b) $2(\log 2))^2$
$
\text { Hint: } \begin{aligned}
\lim _{x \rightarrow 0} \frac{8^x-4^x-2^x+1^x}{x^2} & =\lim _{x \rightarrow 0} \frac{4^x\left(2^x-1\right)-\left(2^x-1\right)}{x^2} \\
& =\lim _{x \rightarrow 0} \frac{\left(4^x-1\right)-\left(2^x-1\right)}{x^2} \\
& =\lim _{x \rightarrow 0} \frac{\left(4^x-1\right)}{x} \frac{\left(2^x-1\right)}{x}=(\log 4) \log 2 \\
& =\log 2^2 \log 2 \\
& =2 \log 2 \cdot \log 2 \\
& =2(\log 2)^2
\end{aligned}
$
Question 9.
If $\mathrm{f}(\mathrm{x})=x(-1)^{\left\lfloor\frac{1}{x}\right\rfloor}, \mathrm{x} \leq \theta$, then the value of $x \rightarrow 0$ is equal to
(a) -1
(b) 0
(c) 2
(d) 4
Solution:
(b) 0
$
\text { Hint: } \begin{aligned}
\lim _{x \rightarrow 0} f(x) & =\lim _{x \rightarrow 0} x(-1)^{\left[\frac{1}{x}\right]}(x \leq 0)=\left[\frac{1}{x}\right]=-1 \\
& =\lim _{x \rightarrow 0} x(-1)^{-1} \\
& =\lim _{x \rightarrow 0}(-x)=0
\end{aligned}
$

Question 10 .
$
\lim _{x \rightarrow 3}\lfloor x\rfloor=
$
(a) 2
(b) 3
(c) does not exist
(d) 0
Solution:
(c) does not exist
Hint:
$
\lim _{x \rightarrow 3}\lfloor x\rfloor=\lim _{x \rightarrow 3}\lfloor x\rfloor=2, \lim _{x \rightarrow 3^{+}}\lfloor x\rfloor=3
$
Limit does not exist

Question 11.
Let the function $\mathrm{f}$ be defined $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{cl}3 x & 0 \leq x \leq 1 \\ -3 x+5 & 1 (a) $\lim _{x \rightarrow 1} f(x)=1$
(b) $\lim _{x \rightarrow 1} f(x)=3$
(c) $\lim _{x \rightarrow 1} f(x)=2$
(d) $\lim _{x \rightarrow 1} f(x)$ does not exist
Solution:

(c) $\lim _{x \rightarrow 1} f(x)=2$
Hint:
$
\begin{aligned}
& \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}} 3 x=3 \\
& \lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}}-3 x+5=-3+5=2
\end{aligned}
$
Limit does not exist

Question 12.
If $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ is defined by $\mathrm{f}(\mathrm{x})=\lfloor x-3\rfloor+|x-4|$ for $\mathrm{x} \in \mathrm{R}$, then $\lim _{x \rightarrow 3^{-}} f(x)$ is equal to
(a) -2
(b) -1
(c) 0
(d) 1
Solution:
(c) 0
Hint:
$
\begin{aligned}
f(x) & =\lfloor x-3\rfloor+|x-4| \text { If } x<0,[x]=n-1 \\
\lim _{x \rightarrow 3^{-}} f(x) & =\lim _{x \rightarrow 3^{-}}[x-3]+|x-4|=\lim _{x \rightarrow 3^{-}}[x-3]+|x-4| \\
& =-1+1 \\
& =0
\end{aligned}
$
Question 13.
$\lim _{x \rightarrow 0} \frac{x e^x-\sin x}{x}$ is
(a) 1
(b) 2
(c) 3
(d) 0
Solution:

(d) 0
$
\text { Hint: } \begin{aligned}
\lim _{x \rightarrow 0} \frac{x e^x-\sin x}{x} & =\lim _{x \rightarrow 0}\left(\frac{x e^x}{x}-\frac{\sin x}{x}\right) \\
& =\lim _{x \rightarrow 0} e^x-\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right) \\
& =1-1 \\
& =0
\end{aligned}
$
Question 14.
$\lim _{x \rightarrow 0} \frac{\sin p x}{\tan 3 x}=4$, then the value of $p$ is
(a) 6
(b) 9
(c) 12
(d) 4
Solution:
(c) 12
$
\begin{gathered}
\text { Hint: } \lim _{x \rightarrow 0} \frac{\frac{\sin p x}{p x} \times p x}{\frac{\tan 3 x}{3 x} \times 3 x}=4 \Rightarrow \frac{\lim _{x \rightarrow 0}\left(\frac{\sin p x}{p x}\right) \times p}{\lim _{x \rightarrow 0}\left(\frac{\tan 3 x}{3 x}\right) \times 3}=4 \\
\frac{p}{3}=4 \Rightarrow p=12
\end{gathered}
$
Question 15.
$\lim _{\alpha \rightarrow \pi / 4} \frac{\sin \alpha-\cos \alpha}{\alpha-\pi / 4}$ is
(a) $\sqrt{2}$
(b) $\frac{1}{\sqrt{2}}$
(c) 1
(d) 2
Solution:

(a) $\sqrt{2}$
Question 16.
$
\lim _{n \rightarrow \infty}\left(\frac{1}{n^2}+\frac{2}{n^2}+\frac{3}{n^2}+\frac{4}{n^2}+\ldots . .+\frac{n}{n^2}\right)
$
(a) $\frac{1}{2}$
(b) 0
(c) 1
(d) $\infty$
Solution:
(a) $\frac{1}{2}$
$
\text { Hint: } \begin{aligned}
& \lim _{n \rightarrow \infty}\left(\frac{1}{n^2}+\frac{2}{n^2}+\frac{3}{n^2}+\frac{4}{n^2}+\ldots .+\frac{n}{n^2}\right) \\
= & \lim _{n \rightarrow \infty}\left(\frac{1+2+3+4+\ldots .+n}{n^2}\right)=\lim _{n \rightarrow \infty} \frac{n(n+1)}{2 n^2}=\lim _{n \rightarrow \infty} \frac{n^2\left(1+\frac{1}{n}\right)}{2 n^2} \\
= & 1 / 2
\end{aligned}
$
Question 17.
$
\lim _{x \rightarrow 0} \frac{e^{\sin x}-1}{x}=
$
(a) 1
(b) $\mathrm{e}$
(c) $\frac{1}{e}$
(d) 0
Solution:

(a) 1
Hint:
$
\begin{aligned}
\lim _{x \rightarrow 0} \frac{e^{\sin x}-1}{x} & =\lim _{x \rightarrow 0} \frac{e^{\sin x}-1}{\sin x} \times \frac{\sin x}{x} \\
& =\lim _{x \rightarrow 0} \frac{e^{\sin x}-1}{\sin x} \times \lim _{x \rightarrow \infty} \frac{\sin x}{x} \Rightarrow 1 \times 1=1
\end{aligned}
$
Question 18.
$
\lim _{x \rightarrow 0} \frac{e^{\tan x}-e^x}{\tan x-x}
$
(a) 1
(b) e
(c) $\frac{1}{2}$
(d) 0
Solution:
(a) 1
Hint:
$
\begin{aligned}
\lim _{x \rightarrow 0} \frac{e^{\tan x}-e^x}{\tan x-x} & \left.=\lim _{x \rightarrow 0} \frac{e^x\left(\frac{e^{\tan x}}{e^x}-1\right)}{(\tan x-x)}\right) \\
& =\lim _{x \rightarrow 0} \frac{e^x\left(e^{\tan x-x}-1\right)}{(\tan x-x)} \\
& =\lim _{x \rightarrow 0} e^x \cdot \lim _{x \rightarrow 0}\left(\frac{e^{\tan x-x}-1}{\tan x-x}\right) \\
& =1 \times 1 \\
& =1
\end{aligned}
$

Question 19.
The value of $\lim _{x \rightarrow 0} \frac{\sin x}{\sqrt{x^2}}$
(a) 1
(b) -1
(c) 0
(d) $\infty$
Solution:
(d) $\infty$
Hint:

$
\begin{aligned}
& =\lim _{x \rightarrow 0} \frac{\sin x}{|x|} \\
& =\lim _{x \rightarrow 0^{-}} \frac{\sin x}{-x}=-1 \\
& =\lim _{x \rightarrow 0^{+}} \frac{\sin x}{x}=1
\end{aligned}
$
So limit does not exist
Question 20.
The value of $\lim _{x \rightarrow k^{-}} x-\lfloor\boldsymbol{x}\rfloor$, , where $\mathrm{k}$ is an integer is
(a) -1
(b) 1
(c) 0
(d) 2
Solution:
(b) 1
$
\text { Hint: } \quad \begin{aligned}
\quad \lim _{x \rightarrow k^{-}} x-\lfloor x\rfloor & =\lim _{x \rightarrow k^{-}} x-\lfloor x\rfloor=k-(k-1) \\
& =k-k x+1=1 .
\end{aligned}
$

Question 21.
At $\mathrm{x}=\frac{3}{2}$ the function $\mathrm{f}(\mathrm{x})=\frac{|2 x-3|}{2 x-3}$ is
(a) Continuous
(b) discontinuous
(c) Differentiate
(d) non-zero
Solution:
(b) discontinuous
$
\begin{aligned}
& \text { Hint: } \quad \lim _{x \rightarrow 3 / 2^{-}} f(x)=\lim _{x \rightarrow 3 / 2^{-}} \frac{|2 x-3|}{2 x-3}=1 \\
& \lim _{x \rightarrow 3 / 2^{+}} f(x)=\lim _{x \rightarrow 3 / 2^{-}} \frac{|2 x-3|}{2 x-3}=7 \\
& \lim _{x \rightarrow 3 / 2^{-}} f(x) \neq \lim _{x \rightarrow 3 / 2^{+}} f(x) \text { is not continuous } \\
&
\end{aligned}
$

Question 22.

Let $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$ be defined by $f(\mathrm{x})=\left\{\begin{array}{ll}x & x \text { is irrational } \\ 1-x & x \text { is rational }\end{array}\right.$ then $\mathrm{f}$ is

(a) Discontinuous at $x=\frac{1}{2}$
(b) Continuous at $x=\frac{1}{2}$
(c) Continuous everywhere
(d) Discontinuous everywhere
Solution:
(b) Continuous at $x=\frac{1}{2}$
Hint:
$
\begin{gathered}
\lim _{x \rightarrow 1 / 2} f(x)=\lim _{x \rightarrow 1 / 2^{-}}(1-x)=1-1 / 2=1 / 2 \\
\lim _{x \rightarrow 1 / 2} f(x)=\lim _{x \rightarrow 1 / 2}(1-x)=1-1 / 2=1 / 2
\end{gathered}
$
$\therefore f(x)$ continuous at $x=1 / 2$

Question 23.
The function $\mathrm{f}(\mathrm{x})=\left\{\begin{array}{ll}\frac{x^2-1}{x^3+1} & x \neq 1 \\ P & x=-1\end{array}\right.$ is not defined for $\mathrm{x}=-1$. The value of $\mathrm{f}(-1)$ so that the function extended by this value is continuous is .............
(a) $\frac{2}{3}$
(b) $-\frac{2}{3}$
(c) 1
(d) 0
Solution:
(b) $-\frac{2}{3}$
Hint: For the function to be continuous at $\mathrm{x}=1$
$
\begin{aligned}
& \lim _{x \rightarrow-1} f(x)=f(-1) \\
& \lim _{x \rightarrow-1} \frac{x^2-1}{x^3+1}=p \\
& \lim _{x \rightarrow-1} \frac{(x-1)(x+1)}{(x+1)\left(x^2-x+1\right)}=p \\
& \frac{-2}{1+1+1}=p \Rightarrow p=-2 / 3
\end{aligned}
$

Question 24.
Let $f$ be a continuous function on $[2,5]$. If $f$ takes only rational values for all $x$ and $f(3)=12$, then $f(4.5)$ is equal to
(a) $\frac{f(3)+f(4.5)}{7.5}$
(b) 12
(c) 17.5
(d) $\frac{f(4.5)-f(3)}{1.5}$
Solution:
(b) 12
Question 25 .
Let a function $\mathrm{f}$ be defined by $f(x)=\frac{x-|x|}{x}$ for $\mathrm{x} \neq 0$ and $f(0)=2$. Then $\mathrm{f}$ is
(a) Continuous nowhere
(b) Continuous everywhere
(c) Continuous for all $x$ except $x=1$
(d) Continuous for all $x$ except $x=0$
Solution:
(d) Continuous for all $x$ except $x=0$
Hint:
$
\begin{aligned}
& f(x)=\left\{\begin{array}{cc}
\frac{x-|x|}{x} & x \neq 0 \\
2 & x=0
\end{array}\right. \\
& \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \frac{x-(-x)}{x}=\lim _{x \rightarrow 0^{-}} \frac{2 x}{x}=2 \\
& \lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} \frac{x-x}{x}=\lim _{x \rightarrow 0^{+}}=0 . \\
& \lim _{x \rightarrow 0^{-}} f(x) \neq \lim _{x \rightarrow 0^{+}} f(x) \\
& \therefore \mathrm{f}(\mathrm{x}) \text { is not continuous at } \mathrm{x}=0 \\
& \Rightarrow \mathrm{f}(\mathrm{x}) \text { is continuous for all except } \mathrm{x}=0
\end{aligned}
$